Given three numbers N, K, and B, the task is to check if N contains only K as digits in Base B.
Examples:
Input: N = 13, B = 3, K = 1
Output: Yes
Explanation:
13 base 3 is 111 which contain all one’s(K).Input: N = 5, B = 2, K = 1
Output: No
Explanation:
5 base 2 is 101 which doesn’t contains all one’s (K).
Naive Approach: A simple solution is to convert the given number N to base B and one by one check if all its digits are K or not.
Time Complexity: O(D), where D is the number of digits in number N
Auxiliary Space: O(1)
Efficient Approach: The key observation in the problem is that any number with all digits as K in base B can be represented as:
These terms are in the form of the Geometric Progression with the first term as K and the common ratio as B.
Sum of G.P. Series:
Therefore, the number in base B with all digits as K is:
Hence, just check if this sum equals N or not. If it’s equal then print “Yes” otherwise print “No”.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include<bits/stdc++.h>using namespace std;// Function to print the number of digits int findNumberOfDigits(int n, int base){ // Calculate log using base change // property and then take its floor // and then add 1 int dig = (floor(log(n) / log(base)) + 1); // Return the output return (dig); }// Function that returns true if n contains // all one's in base b int isAllKs(int n, int b, int k){ int len = findNumberOfDigits(n, b); // Calculate the sum int sum = k * (1 - pow(b, len)) / (1 - b); if(sum == n) { return(sum); }}// Driver code int main(){ // Given number N int N = 13; // Given base B int B = 3; // Given digit K int K = 1; // Function call if (isAllKs(N, B, K)) { cout << "Yes"; } else { cout << "No"; }}// This code is contributed by vikas_g |
C
// C implementation of the approach#include <stdio.h>#include <math.h>// Function to print the number of digits int findNumberOfDigits(int n, int base){ // Calculate log using base change // property and then take its floor // and then add 1 int dig = (floor(log(n) / log(base)) + 1); // Return the output return (dig); }// Function that returns true if n contains // all one's in base b int isAllKs(int n, int b, int k){ int len = findNumberOfDigits(n, b); // Calculate the sum int sum = k * (1 - pow(b, len)) / (1 - b); if(sum == n) { return(sum); }}// Driver code int main(void){ // Given number N int N = 13; // Given base B int B = 3; // Given digit K int K = 1; // Function call if (isAllKs(N, B, K)) { printf("Yes"); } else { printf("No"); } return 0;}// This code is contributed by vikas_g |
Java
// Java implementation of above approachimport java.util.*;class GFG{// Function to print the number of digitsstatic int findNumberOfDigits(int n, int base){ // Calculate log using base change // property and then take its floor // and then add 1 int dig = ((int)Math.floor(Math.log(n) / Math.log(base)) + 1); // Return the output return dig;}// Function that returns true if n contains// all one's in base bstatic boolean isAllKs(int n, int b, int k){ int len = findNumberOfDigits(n, b); // Calculate the sum int sum = k * (1 - (int)Math.pow(b, len)) / (1 - b); return sum == n;}// Driver codepublic static void main(String[] args){ // Given number N int N = 13; // Given base B int B = 3; // Given digit K int K = 1; // Function call if (isAllKs(N, B, K)) System.out.println("Yes"); else System.out.println("No");}}// This code is contributed by offbeat |
Python3
# Python3 program for the above approachimport math# Function to print the number of digitsdef findNumberOfDigits(n, base): # Calculate log using base change # property and then take its floor # and then add 1 dig = (math.floor(math.log(n) / math.log(base)) + 1) # Return the output return dig# Function that returns true if n contains# all one's in base bdef isAllKs(n, b, k): len = findNumberOfDigits(n, b) # Calculate the sum sum = k * (1 - pow(b, len)) / (1 - b) return sum == N# Driver code# Given number NN = 13# Given base BB = 3# Given digit KK = 1# Function callif (isAllKs(N, B, K)): print("Yes")else: print("No") |
C#
// C# implementation of above approachusing System;class GFG{ // Function to print the number of digits static int findNumberOfDigits(int n, int bas) { // Calculate log using base change // property and then take its floor // and then add 1 int dig = ((int)Math.Floor(Math.Log(n) / Math.Log(bas)) + 1); // Return the output return dig;} // Function that returns true if n contains // all one's in base b static bool isAllKs(int n, int b, int k) { int len = findNumberOfDigits(n, b); // Calculate the sum int sum = k * (1 - (int)Math.Pow(b, len)) / (1 - b); return sum == n; } // Driver codepublic static void Main(){ // Given number N int N = 13; // Given base B int B = 3; // Given digit K int K = 1; // Function call if (isAllKs(N, B, K)) Console.Write("Yes"); else Console.Write("No"); }}// This code is contributed by vikas_g |
Javascript
<script>// Javascript implementation of the approach // Function to print the number of digits function findNumberOfDigits(n, base){ // Calculate log using base change // property and then take its floor // and then add 1 var dig = (Math.floor(Math.log(n) / Math.log(base)) + 1); // Return the output return (dig); }// Function that returns true if n contains // all one's in base b function isAllKs(n, b, k){ var len = findNumberOfDigits(n, b); // Calculate the sum var sum = k * (1 - Math.pow(b, len)) / (1 - b); if(sum == n) { return(sum); }}// Driver code // Given number N var N = 13;// Given base B var B = 3;// Given digit K var K = 1;// Function call if (isAllKs(N, B, K)){ document.write( "Yes");} else{ document.write("No");}// This code is contributed by rrrtnx.</script> |
Output:
Yes
Time Complexity: O(log(D)), where D is the number of digits in number N
Auxiliary Space: O(1)
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